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Question
If n is a positive integer greater than 1, then
a− nC1(a−1)+ nC2(a−2)−⋯+(−1)n(a−n) is equal to
If n is a positive integer greater than 1, then
a− nC1(a−1)+ nC2(a−2)−⋯+(−1)n(a−n) is equal to
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Solution
The correct option is C 0
We have, tr+1=(−1)r⋅ nCr⋅(a−r)
=(−1)r(a⋅nCr−r⋅nCr)
=(−1)r(a⋅nCr−n⋅n−1Cr−1)
(∵ r⋅ nCr=n⋅ n−1Cr−1)
Putting r=0, 1, 2,⋯, n, we get
t1=a⋅ nC0−n⋅0
t2=−(a⋅ nC1−n⋅ n−1C0)
t3=a⋅ nC2−n⋅ n−1C1
⋯ ⋯ ⋯
tn+1=(−1)n(a⋅nCn−n⋅n−1Cn−1)
Adding,
Sum=a[nC0− nC1+ nC2− nC3+⋯+(−1)n⋅ nCn] +n[n−1C0− n−1C1+ n−1C2−⋯+(−1)n⋅ n−1Cn−1]
=a×0+n×0=0
We have, tr+1=(−1)r⋅ nCr⋅(a−r)
=(−1)r(a⋅nCr−r⋅nCr)
=(−1)r(a⋅nCr−n⋅n−1Cr−1)
(∵ r⋅ nCr=n⋅ n−1Cr−1)
Putting r=0, 1, 2,⋯, n, we get
t1=a⋅ nC0−n⋅0
t2=−(a⋅ nC1−n⋅ n−1C0)
t3=a⋅ nC2−n⋅ n−1C1
⋯ ⋯ ⋯
tn+1=(−1)n(a⋅nCn−n⋅n−1Cn−1)
Adding,
Sum=a[nC0− nC1+ nC2− nC3+⋯+(−1)n⋅ nCn] +n[n−1C0− n−1C1+ n−1C2−⋯+(−1)n⋅ n−1Cn−1]
=a×0+n×0=0
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